On Wed, Nov 21, 2012 at 1:34 AM, Virgil Stokes <vs@it.uu.se> wrote:
> But, again, it is an issue for the algorithm given in Table 3 (p. 2248
> of paper). Look at step 8. and equation (30).
>
I see no Table 3, nor page 2248 in the paper you attached. Which paper are
you talking about?
> As stated in this step "The square-root of the filtered state-error
> covariance" is returned as [image: $T_{22}$] from step 5. where a QR
> decomposition is performed for the triangularization. The [image:
> $S_{k+1|k+1}$] matrix must have diagonal elements > 0 (I leave this to
> you to think about). When I perform step 5. with MATLAB, I *always* get a [image:
> $T_{22}$] that has diagonal elements > 0 for my application (which is
> satisfying). When I perform step 5. with numpy, for the same matrix on the
> RHS of (27) I *do not always* get diagonal elements > 0 for [image:
> $T_{22}$] --- herein lies the problem that I have been trying to explain.
>>Anyway, I would say that an algorithm that depend on the sign of the
entries of R is not correct.
Alejandro.
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